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Power on slopes and with towing

Edexcel IAL Mathematics · WME02 Mechanics 2

On a slope the weight has a component along the road, which the engine must also overcome when going uphill (and which helps when going downhill). Towing problems add a second body joined by a tow bar.

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Up and down slopes

Resolving along a slope inclined at α:

At maximum speed (or constant speed), a = 0. Slopes are often given as sin α = 1/20 ("1 in 20").

Towing

A car towing a trailer by a light horizontal tow bar: use the whole system to find the acceleration (the tension is internal), then the trailer alone to find the tension. If the trailer slows the car, the tow bar is in tension; if it pushes (braking downhill), it is in thrust.

Work-energy with an engine

Over a distance, the work done by the engine = increase in KE + increase in PE + work done against resistance. If the power is constant over a time t, the engine's work is Pt.

Worked example

A car of mass 1200 kg moves up a slope with sin α = 1/20 at a constant 15 m/s against a resistance of 600 N. Find the power of its engine.

  1. Constant speed: F = 600 + 1200 × 9.8 × (1/20) = 600 + 588 = 1188 N.
  2. P = 1188 × 15 = 17800 W.

Answer: 17.8 kW

Key idea: Uphill: P/v = R + mg sin α + ma. Downhill: P/v + mg sin α = R + ma. Towing: whole system for a, trailer alone for T.

Check yourself

  1. A car of mass 1000 kg works at 30 kW up a slope with sin α = 1/14 against a resistance of 500 N. Find its maximum speed.
    Show answer

    25 m/s. F = 500 + 1000 × 9.8/14 = 500 + 700 = 1200 N. v = 30 000/1200 = 25 m/s.

  2. A car of mass 1000 kg works at 12 kW down a slope with sin α = 1/49 against a resistance of 800 N. Find its maximum speed.
    Show answer

    20 m/s. F + 200 = 800, so F = 600 N. v = 12 000/600 = 20 m/s.

  3. A car of mass 900 kg works at 20 kW up a slope with sin α = 1/15 against a resistance of 400 N. Find its acceleration at 10 m/s, to 3 significant figures.
    Show answer

    1.12 m/s2. F = 2000 N. 2000 - 400 - 588 = 900a, a = 1.12 m/s2.

  4. A car of mass 1000 kg tows a trailer of mass 500 kg on a level road. Resistances are 400 N on the car and 200 N on the trailer. The engine works at 24 kW. Find the tension in the tow bar when the speed is 12 m/s, to 3 significant figures.
    Show answer

    667 N. Whole system: 2000 - 600 = 1500a, a = 0.933 m/s2. Trailer: T - 200 = 500a, so T = 667 N.

  5. A car of mass 1000 kg moves 200 m up a slope with sin α = 0.1, speeding up from 10 m/s to 20 m/s against a constant resistance of 500 N. Find the work done by the engine, in kJ, to 3 significant figures.
    Show answer

    446 kJ. KE gain 150 000 J, PE gain 1000 × 9.8 × 20 = 196 000 J, work against resistance 100 000 J. Total 446 kJ.

  6. A lorry climbs a hill at constant speed. Which statement is correct?
    • Driving force = resistance
    • Driving force = resistance + mg sin α
    • Driving force = mg sin α - resistance
    • Driving force = 0
    Show answer

    Driving force = resistance + mg sin α. No acceleration, so forces along the slope balance: the engine must overcome both.

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Last updated 8 November 2026. Spotted a mistake? Tell us.